A light rod of length 1 m is pivoted at its centre, with two masses of 5 kg and 2 kg suspended from its ends, as shown in the figure. Determine the initial angular acceleration of the rod, assuming it is horizontal at the beginning.

Step-by-step solution

Let the length of the rod be L = 1 m . Since the rod is pivoted at its centre, the distance of each mass from the pivot is r = L 2 = 0.5 m . The forces acting on the system are the weights of the two masses. The net torque about the pivot is the difference between the torques produced by the 5 kg and 2 kg masses: = m₁ g r - m₂ g r = (m₁ - m₂) g r Substituting the given values (taking g = 9.8 m/s ^2 ): = (5 - 2) 9.8 0.5 = 3 4.9 = 14.7 N m The total moment of inertia I of the system about the pivot, assuming the rod